一个流缓冲的内容复制到一个字符串 [英] Copy a streambuf's contents to a string

问题描述

显然提高:: ASIO :: async_read 不喜欢字符串，如提振唯一的过载:: ASIO ::缓​​冲允许我创建 const_buffer S，所以我坚持与阅读一切都变成流缓冲。

现在我想的streambuf的内容复制到一个字符串，但它显然只支持写入为char *（ sgetn（）），创建具有流缓冲的istream和使用函数getline（）。

Apparently boost::asio::async_read doesn't like strings, as the only overload of boost::asio::buffer allows me to create const_buffers, so I'm stuck with reading everything into a streambuf.
Now I want to copy the contents of the streambuf into a string, but it apparently only supports writing to char* (sgetn()), creating an istream with the streambuf and using getline().

是否有任何其他的方法来创建具有的streambuf内容的字符串没有过多的复制？

Is there any other way to create a string with the streambufs contents without excessive copying?

推荐答案

我不知道这是否算作过度复制的，但您可以使用一个字符串流：

I don't know whether it counts as "excessive copying", but you can use a stringstream:

std::ostringstream ss;
ss << someStreamBuf;
std::string s = ss.str();

像，读一切从标准输入到一个字符串，执行

Like, to read everything from stdin into a string, do

std::ostringstream ss;
ss << std::cin.rdbuf();
std::string s = ss.str();

另外，您还可以使用 istreambuf_iterator 。你将不得不衡量这或以上的方式是更快的 - 我不知道

Alternatively, you may also use a istreambuf_iterator. You will have to measure whether this or the above way is faster - i don't know.

std::string s((istreambuf_iterator<char>(someStreamBuf)), 
               istreambuf_iterator<char>());

注意 someStreamBuf 以上是为了重新present一个流缓冲* ，所以取它的地址适当。还要注意身边在最后一个例子，第一个参数的额外括号，所以它不跨preT它作为一个函数声明返回一个字符串，并考虑迭代器和另一个函数指针（最让人头疼的解析）。

Note that someStreamBuf above is meant to represent a streambuf*, so take its address as appropriate. Also note the additional parentheses around the first argument in the last example, so that it doesn't interpret it as a function declaration returning a string and taking an iterator and another function pointer ("most vexing parse").

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